What is the integral of (cos(pi/(x^{39)}))/(x^{40)} ?

The integral of (cos(pi/(x^{39)}))/(x^{40)} is -1/(39pi)sin(pi/(x^{39)})+C

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integral of (cos(pi/(x^{39)}))/(x^{40)}

\int \frac{cos ( \frac{π}{x ^{39}} )}{x ^{40}} d x

- \frac{1}{39 π} sin (\frac{π}{x ^{39}}) + C

Solution steps

\int \frac{cos ( \frac{π}{x ^{39}} )}{x ^{40}} d x

Apply u-substitution

: \int \frac{cos ( \frac{π}{u} )}{39 u ^{2}} d u

= \int \frac{cos ( \frac{π}{u} )}{39 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{39} \cdot \int \frac{cos ( \frac{π}{u} )}{u ^{2}} d u

Apply u-substitution

: \int - \frac{cos ( v )}{π} d v

= \frac{1}{39} \cdot \int - \frac{cos ( v )}{π} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{39} (- \frac{1}{π} \cdot \int cos (v) d v)

Use the common integral:

\int cos (v) d v = sin (v)

= \frac{1}{39} (- \frac{1}{π} sin (v))

Substitute back

= \frac{1}{39} (- \frac{1}{π} sin (\frac{π}{x ^{39}}))

Simplify

\frac{1}{39} (- \frac{1}{π} sin (\frac{π}{x ^{39}})) : - \frac{1}{39 π} sin (\frac{π}{x ^{39}})

= - \frac{1}{39 π} sin (\frac{π}{x ^{39}})

Add a constant to the solution

= - \frac{1}{39 π} sin (\frac{π}{x ^{39}}) + C

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What is the integral of (cos(pi/(x^{39)}))/(x^{40)} ?
The integral of (cos(pi/(x^{39)}))/(x^{40)} is -1/(39pi)sin(pi/(x^{39)})+C